Python Program to Count trailing zeroes in factorial of a number

Given an integer n, write a function that returns the count of trailing zeroes in n!  

Examples :

Input: n = 5
Output: 1 
Factorial of 5 is 120 which has one trailing 0.

Input: n = 20
Output: 4
Factorial of 20 is 2432902008176640000 which has
4 trailing zeroes.

Input: n = 100
Output: 24

Trailing 0s in n! = Count of 5s in prime factors of n!
                  = floor(n/5) + floor(n/25) + floor(n/125) + ....

Method 1:

# Python3 program to
# count trailing 0s
# in n !

# Function to return
# trailing 0s in
# factorial of n


def findTrailingZeros(n):

    # Initialize result
    count = 0

    # Keep dividing n by
    # powers of 5 and
    # update Count
    i = 5
    while (n / i >= 1):
        count += int(n / i)
        i *= 5

    return int(count)


# Driver program
n = 100
print("Count of trailing 0s " +
      "in 100 ! is", findTrailingZeros(n))

Count of trailing 0s in 100 ! is 24

Time Complexity:  O(log₅n)

Auxiliary Space: O(1)

Please refer complete article on Count trailing zeroes in factorial of a number for more details!

Method 2: Using math.factorial() and for loop

# Python3 program to
# count trailing 0s
# in n !

# Function to return
# trailing 0s in
# factorial of n


def findTrailingZeros(n):
    import math
    c = 0
    x = math.factorial(n)
    s = str(x)
    a = s[::-1]
    for i in a:
        if(i != "0"):
            break
        else:
            c += 1
    return c


# Driver program
n = 100
print("Count of trailing 0s " +
      "in 100 ! is", findTrailingZeros(n))

Count of trailing 0s in 100 ! is 24

Time complexity: O(n)
Space complexity: O(n)

Method 3: Using recursion

Follow the steps below:

• The function countTrailingZeroes takes the parameter n and a count variable count which is initialized to 0.
• If n is equal to 0, the function returns the count variable. This is because there are no more numbers to check for multiples of 5.
• If n is not equal to 0, the function updates the count variable by dividing n by 5 and adding the result to the count variable.
• The function then calls itself, this time with n divided by 5 and the updated count variable as parameters.
• The main function initializes the n variable and calls the countTrailingZeroes function, storing the result in the result variable.
• Finally, the main function prints out the number of trailing zeros in n!.

def countTrailingZeroes(n, count = 0):
    if n == 0:
        return count
    else:
        count += n//5
        return countTrailingZeroes(n//5, count)

def main():
    n = 100
    result = countTrailingZeroes(n)
    print("The number of trailing zeros in %d! is %d" % (n, result))

if __name__ == '__main__':
    main()

The number of trailing zeros in 100! is 24

Time complexity: O(log n). This is because in each recursive call, the value of n is divided by 5 until it becomes 0. Hence, the number of recursive calls is proportional to log n to the base 5.
Auxiliary space: O(log n) as well, because of the number of function calls required for the recursive function. 

Method 4: Using Iterative method

Use a while loop to count the number of multiples of 5, starting from 5, and then multiplies the current count by 5 to count the next power of 5. The loop continues until the current power of 5 is greater than n. Finally, the function returns the total count of powers of 5, which corresponds to the number of trailing zeros in n!.

import math

def findTrailingZeros(n):
    # count powers of 5 in prime factorization
    trailing_zeros = 0
    i = 5
    while i <= n:
        trailing_zeros += n // i
        i *= 5
    
    return trailing_zeros

# Driver program
n = 100
print("Count of trailing 0s " +
      "in 100 ! is", findTrailingZeros(n))

Count of trailing 0s in 100 ! is 24

Time complexity: O(log n)
Auxiliary space: O(1)

kartik

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